If P=_0^/2^3/2x^3/2x+^3/2x,dx (so P=/4 by King's rule), then _0^2P(x x x)/(^4x+^4x),dx equ…

Question

If $P=\displaystyle\int_0^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx$ (so $P=\pi/4$ by King's rule), then $\displaystyle\int_0^{2P}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\,dx$ equals:

Accepted Answer

By King's rule, $2P=\displaystyle\int_0^{\pi/2}dx = \pi/2$, so $P=\pi/4$ and $2P=\pi/2$.

$$I = \int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\,dx$$

By King's rule ($x\to\pi/2-x$) applied to $I$ itself, $2I = \dfrac\pi2\displaystyle\int_0^{\pi/2}\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}dx$. With $\tan^2x=u$:

$$\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx = \frac12\int_0^\infty\frac{du}{u^2+1} = \frac{\pi}{4}$$

So $2I = \dfrac\pi2\cdot\dfrac\pi4 = \dfrac{\pi^2}8 \implies I=\dfrac{\pi^2}{16}$.

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